What is the solution for y^{''}-4y=6cos(3t+30) ?

The solution for y^{''}-4y=6cos(3t+30) is y=c_{1}e^{2t}+c_{2}e^{-2t}-6/13 cos(3t+30)

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y^{''}-4y=6cos(3t+30)

y^{^{''}} - 4 y = 6 cos (3 t + 3 0^{\circ})

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{6}{13} cos (3 t + 3 0^{\circ})

Solution steps

y^{''} (t) - 4 y = 6 cos (3 t + 3 0^{\circ})

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{6}{13} cos (3 t + 3 0^{\circ})

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{6}{13} cos (3 t + 3 0^{\circ})

y^{^{''}} - 7 y^{^{'}} + 6 y = 3 e^{t}

y^{^{''}} + 4 y = t e^{3 t} + 3 sin (4 t)

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{x}

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} - 2 x

y^{^{''}} + 25 y = 11 sec^{2} (5 t)

What is the solution for y^{''}-4y=6cos(3t+30) ?
The solution for y^{''}-4y=6cos(3t+30) is y=c_{1}e^{2t}+c_{2}e^{-2t}-6/13 cos(3t+30)